What is the answer to \sqrt[3]{11-x}+sqrt(25+x)=6 ?

The answer to \sqrt[3]{11-x}+sqrt(25+x)=6 is x=11,x=-16,x=75

Solve for x: \sqrt[3]{11-x}+sqrt(25+x)=6

The solution is x=11,x=-16,x=75

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x = 11, x = - 16, x = 75

Solution steps

Rewrite the equation with

and

x = - u^{3} + 11

u + 25 + (- u^{3} + 11) = 6

Solve

u + 25 + (- u^{3} + 11) = 6 : u = 0, u = 3, u = - 4

u = 0, u = 3, u = - 4

Substitute back

solve for

x

Solve

x = 11, x = - 16, x = 75

Verify Solutions:

x = 11

True

, x = - 16

True

, x = 75

True

The solutions are

x = 11, x = - 16, x = 75

x^{2} + 48 x = 10

(x - 2)^{\frac{1}{3}} = - 5

2 = x^{- \frac{1}{2}}

4 x + 5 - 2 x + 2 + 1 = 0

\frac{y ^{\frac{3}{2}}}{27} = 1

What is the answer to \sqrt[3]{11-x}+sqrt(25+x)=6 ?
The answer to \sqrt[3]{11-x}+sqrt(25+x)=6 is x=11,x=-16,x=75
Solve for x: \sqrt[3]{11-x}+sqrt(25+x)=6
The solution is x=11,x=-16,x=75